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# A force of $50\,{\text{kgf}}$ is applied to the smaller piston of a hydraulic machine. Neglecting the friction, the force exerted on the large piston, if the diameters of the piston are $5\,{\text{cm}}$ and $25\,{\text{cm}}$ respectively is ‘$X$’ ${\text{N}}$ . Find $\dfrac{X}{{250}}$ .

Last updated date: 28th May 2024
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Hint: First we will calculate the areas of the smaller and the larger pistons. Then we will calculate the pressure applied onto each of them. Eventually, we will equate the two equations and manipulate accordingly to obtain the result.

In the given question, we are supplied with the following data:
The diameter of the smaller piston is $5\,{\text{cm}}$ .
The diameter of the larger piston is $25\,{\text{cm}}$ .
The force exerted on the smaller piston is $50\,{\text{kgf}}$ .
We are asked to find the force exerted on the larger piston.
We will apply the trick that the pressure is evenly distributed in all the pistons, as we know water is an incompressible liquid. The pressure you put on the smaller piston and the pressure on the larger piston will be equal.
We know the formula which relates pressure, force and area is given below:
$P = \dfrac{F}{A}$ …… (1)
Where,
$P$ indicates pressure.
$F$ indicates force applied.
$A$ indicates the area on which the force is applied.
Again, we will find the area of the smaller piston:
$A = \pi {\left( {\dfrac{D}{2}} \right)^2}$ …… (2)
Where,
$A$ indicates area.
$D$ indicates diameter of the piston.
Now, we calculate the area of the smaller piston by substituting the required values in equation (2):
$A = \pi {\left( {\dfrac{D}{2}} \right)^2} \\ A = 3.14 \times {\left( {\dfrac{5}{2}} \right)^2} \\ A = 19.6\,{\text{c}}{{\text{m}}^2} \\$
Therefore, the area of the smaller piston is found to be $19.6\,{\text{c}}{{\text{m}}^2}$ .

So, the pressure can be calculated by using the equation (1), as follows:
$P = \dfrac{F}{A}$
$P = \dfrac{{50\,{\text{kgf}}}}{{19.6\,{\text{c}}{{\text{m}}^2}}}$ …… (3)

Again, we will find the area of the larger piston:
Now, we calculate the area of the smaller piston by substituting the required values in equation (2):
$A = \pi {\left( {\dfrac{D}{2}} \right)^2} \\ A = 3.14 \times {\left( {\dfrac{{25}}{2}} \right)^2} \\ A = 490.6\,{\text{c}}{{\text{m}}^2} \\$
Therefore, the area of the smaller piston is found to be $490.6\,{\text{c}}{{\text{m}}^2}$ .

So, the pressure can be calculated by using the equation (1), as follows:
$P = \dfrac{F}{A}$
$P = \dfrac{X}{{490.6\,{\text{c}}{{\text{m}}^2}}}$ …… (4)

Now, we equate equation (3) and (4), and we get:
$\dfrac{{50\,{\text{kgf}}}}{{19.6\,{\text{c}}{{\text{m}}^2}}} = \dfrac{X}{{490.6\,{\text{c}}{{\text{m}}^2}}} \\ X = \dfrac{{50\,{\text{kgf}} \times 490.6\,{\text{c}}{{\text{m}}^2}}}{{19.6\,{\text{c}}{{\text{m}}^2}}} \\ X = 1251.51\,{\text{kgf}} \\ X \sim 1252\,{\text{kgf}} \\$
Hence, the force exerted on the larger piston is $1252\,{\text{kgf}}$ .

Note:The given problem is based on the mechanical properties of fluid. While calculating the area, you can also use the S.I units, there’s no issue and the answer will come the same because the units of the area will eventually cancel each other. These types of hydraulic lifts are used to lift heavy vehicles, by applying pressure on one end.